$$\textrm{H}_2\textrm{CO}_3 \rightleftharpoons \textrm{HCO}_3^{-} + \textrm{H}^+, \; \textrm{pK}_2 = 3.6, \; K_2 = 10^{-3.6}.$$ Thus, at pH = 7.4, the majority of the phosphate will be in the HPO$_3^{2-}$ form, and nearly 100% of the bicarbonate will be in the unprotonated HCO$_3^-$ form.

Although the equilibrium constant for this reaction is very low, the relatively small amount of H$_2$CO$_3$ formed will be associated with a much higher amount of HCO$_3^-$, given the dissociation reaction detailed above. To see how these systems work in the body, consider the following problems.

(a.) At a normal pH of plasma is 7.4, and a total carbon dioxide concentration [$\Sigma$CO$_2$] = [CO$_2$] + [H$_2$CO$_3$] + [HCO$_3^2$] = 30 mM, and assuming that the carbonic anhydrase reaction is in equilibrium, what is the free concentration of carbon dioxide in the plasma?

To solve this problem, express the above sum for total carbon dioxide concentration [$\Sigma$CO$_2$] in terms of the unknown [CO$_2$] and the known pH:

These calculations yield [CO$_2$] = 1.66 mM, meaning about 94\% of the $\Sigma$CO$_2$ is in the form of bicarbonate.

(b.) In Chapter 2 of the Silverthorn textbook [1], Question 22 asks, “What effect does hard work by your muscles have on the pH of the blood?” Here, we examine that question quantitatively in terms of the bicarbonate system. Assume the concentrations from the above problem as an initial condition, with the addition of 1.5 mM total inorganic phosphate in the plasma. How does pH change as additional CO$_2$ is injected into the system?

Your solution will require simultaneously solving three conservation relationships. The first is for total phosphate

Here we have three equations and three unknowns. In MATLAB we can solve systems like this using the function fsolve. The first step is to write a function that computes the three equations as a function of the three variables:

function f = conserve3(x,delta_CO2,Po,SCO2o,Ho,H2PO3o,HCO3o)

% Chemical constants.

K1 = 10^(-6.8);

K2 = 10^(-3.60);

Keq = 2.7e-3;

% The three unknowns

HPO3 = x(1);

CO2 = x(2);

H = x(3);

% Conservation equations (should equal 0)

f(1,:) = Po - HPO3*(1 + H/K1);

f(2,:) = SCO2o + delta_CO2 - CO2*(1 + Keq + K2*Keq/H);

f(3,:) = Ho + H2PO3o - HCO3o - H - Po*(H/K1)/(1+H/K1) + CO2*K2*Keq/H;

Notice that the output of the function, 'f', is a vector that is equal to 0 when Equation (4) is satisfied. In addition to the vector of unknowns, x, the function accepts as inputs the injected CO2, the total phosphate, initial total CO2, initial hydrogen ion activity, initial protonated phosphate, and initial protonated bicarb: 'delta_CO2', 'Po', 'SCO2o', 'Ho', 'H2PO3o', 'HCO3o'.

To call the 'fsolve' function and compute the equilibrium concentration, we use the following syntax

delta_CO2 = +10e-3;

x = fsolve(@conserve3,x,[],delta_CO2,Po,SCO2o,Ho,H2PO3o,HCO3o);

pH = -log10(x(3))

This calculation yields a pH of approximately 6.6, illustrating that injection of CO$_2$ causes a drop in pH. (What happens to the predicted drop in pH if the total phosphate concentration is increased? You should find that increasing the phosphate concentration results in smaller changes in pH for a given amount of CO$_2$ injected.)